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Moldova Team Selection Test
2012 Moldova Team Selection Test
10
4a_1-2b_1=7
4a_1-2b_1=7
Source: Moldova TST 2012
March 9, 2023
number theory
Problem Statement
Let
f
:
R
→
R
,
f
(
x
,
y
)
=
x
2
−
2
y
.
f:\mathbb{R}\rightarrow\mathbb{R}, f(x,y)=x^2-2y.
f
:
R
→
R
,
f
(
x
,
y
)
=
x
2
−
2
y
.
Define the sequences
(
a
n
)
n
≥
1
(a_n)_{n\geq1}
(
a
n
)
n
≥
1
and
(
b
n
)
n
≥
1
(b_n)_{n\geq1}
(
b
n
)
n
≥
1
such that
a
n
+
1
=
f
(
a
n
,
b
n
)
,
b
n
+
1
=
f
(
b
n
,
a
n
)
.
a_{n+1}=f(a_n,b_n), b_{n+1}=f(b_n,a_n).
a
n
+
1
=
f
(
a
n
,
b
n
)
,
b
n
+
1
=
f
(
b
n
,
a
n
)
.
If
4
a
1
−
2
b
1
=
7
:
4a_1-2b_1=7 :
4
a
1
−
2
b
1
=
7
:
a) find the smallest
k
∈
N
k\in\mathbb{N}
k
∈
N
for which the number
p
=
2
k
⋅
(
2
512
a
9
−
b
9
)
p=2^k\cdot(2^{512}a_9-b_9)
p
=
2
k
⋅
(
2
512
a
9
−
b
9
)
is an integer. b) prove that
2
2
10
+
2
2
9
+
1
2^{2^{10}}+2^{2^9}+1
2
2
10
+
2
2
9
+
1
divides
p
.
p.
p
.
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